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KatRina [158]
3 years ago
11

During one cycle, a refrigerator removes 500 J from a cold reservoir and discharges 800 J to its hot reservoir. (a) What is its

coefficient of performance? (b) How much work per cycle does it require to operate?
Physics
1 answer:
Paraphin [41]3 years ago
5 0

Answer:

a)COP = 1.67

b) W = 300 J

Explanation:

given,

refrigerator removes = 500 J from a cold reservoir

discharge = 800 J to its hot reservoir

a) COP = \dfrac{Q}{w}      

   COP = \dfrac{500}{800-500}

   COP = 1.67                        

b) heat per cycle exhausted to the

     W + 500 = 800                        

     W = 800- 500                        

      W = 300 J                                  

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What happens to the gravitational force between two masses when the distance between the masses is doubled?
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Answer:

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In this problem, the distance between the masses is doubled, so

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F'=\frac{Gm_1 m_2}{(2r)^2}=\frac{1}{4}\frac{Gm_1m_2}{r^2})\frac{F}{4}

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To determine the coefficient of static friction between two materials, an engineer places a small sample of one material on a ho
Fudgin [204]

This question is incomplete, the missing image is uploaded along this answer.

Answer:

the coefficient of friction is 0.32

Explanation:

 Given the data in the question;

we make use of kinematic equation of motion;

ω = ω₀ + ∝t

we substitute

ω = ( 0 rad/s ) + ( 0.4 rad/s² )( 9.903 s )

ω = 3.9612 rad/s

The centripetal force acting on the sample is;

Fc = mrω²

from the image; r = 200 mm = 0.2 m

so we substitute

Fc = m(0.2 m ) ( 3.9612 rad/s )²

Fc = (3.13822 m/s²)m

we know that the frictional force between the two materials should be providing the necessary centripetal force to rotate the sample object;

f = Fc

μN = Fc

μmg = (3.13822 m/s²)m

μ = (3.13822 m/s²)m / mg

μ = (3.13822 m/s²) / g

acceleration due to gravity g = 9.8 m/s²

so

μ = (3.13822 m/s²) / 9.8 m/s²

μ = 0.32

Therefore, the coefficient of friction is 0.32

5 0
3 years ago
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